It is not difficult to use multi-variable calculus to show that an inversion map is conformal (e.g., showing that the Jacobian matrix at each point of an inversion map is a scalar multiple of a reflection matrix). Let's say that you have an inversion $\iota$ about the point $(h,k)$ with radius $r$. Then
$$\iota(x,y)=\left(h+\frac{r^2(x-h)}{(x-h)^2+(y-k)^2},k+\frac{r^2(y-k)}{(x-h)^2+(y-k)^2}\right).$$
The Jacobian matrix of $\iota$ is then
$$J(x,y)=\frac{r^2}{(x-h)^2+(y-k)^2}R(x,y)$$
where $R(x,y)$ is the reflection matrix
$$R(x,y)=\begin{bmatrix}-\cos\theta(x,y) &-\sin\theta(x,y)\\ -\sin\theta(x,y)&\cos\theta(x,y)\end{bmatrix}$$
s.t. $$\cos\theta(x,y)=\frac{(x-h)^2-(y-k)^2}{(x-h)^2+(y-k)^2}$$ and $$\sin\theta(x,y)=\frac{2(x-h)(y-k)}{(x-h)^2+(y-k)^2}.$$ Indeed, if a differentiable bijection $\phi:M\to M$ where $M$ is an open subset of $\Bbb R^n$, is such that the Jacobian matrix of $\phi$ at each point $s\in M$ is of the form $\sigma(s)A(s)$, where $\sigma:M\to\Bbb R$ is a strictly positive-valued function and $A(s)$ is an orthogonal matrix for each $s\in M$, then $\phi$ is a conformal map.
Here is a differential geometry argument. I merely intended to give the thread some answer. I do not immediately see an elementary geometry proof, since the discussion of tangents of an arbitrary curve should involve some level of calculus/differential geometry. Please enlighten me if there is a way to avoid calculus/differential geometry completely.
An inversion map $\iota$ can be extended to the Riemann sphere $\hat{\Bbb C}=\Bbb C\cup \{\infty\}$. (Say, $\iota$ is the inversion about a circle centered at some point $o$. Then $\iota(\infty)=o$ and $\iota(o)=\infty$. A straight line is actually a circle in $\hat{\Bbb C}$ that passes through $\infty$, and the inversion w.r.t. this straight line is the same as the reflection about the line.) This map can be easily seen to be smooth, but this is the pivotal part of the problem so an interested reader should check it. Since $\iota$ is an involution, it is a auto-diffeomorphism of $\hat{\Bbb C}$.
Let $I$ be the open interval $(-1,1)$. Suppose that the curves $\alpha,\beta$ which are, respectively, the images of two differentiable embeddings $a,b:I\to\hat{\Bbb C}$ are tangent at a point $p=a(0)=b(0)$, then the tangent spaces $T_p\alpha=da_0(T_0I)$ and $T_p\beta=db_0(T_0I)$ coincide. (Differentiability of $a$ and $b$ is necessary. Otherwise it makes no sense to discuss the term tangents.)
Let $q=\iota(p)$. Then the differential map $d\iota_p:T_p\hat{\Bbb C}\to T_q\hat{\Bbb C}$ is an isomorphism of vector spaces due to $\iota$ being a diffeomorphism of $\hat{\Bbb C}$. Therefore $d\iota_p$ sends
$$T_p\alpha\mapsto T_q\iota(\alpha)$$
and
$$T_p\beta\mapsto T_q\iota(\beta).$$
But $T_p\alpha=T_p\beta$, so
$$T_q\iota(\alpha)=T_q\iota(\beta).$$
Hence the inversion images $\iota(\alpha)$ and $\iota(\beta)$ of $\alpha$ and $\beta$ under $\iota$ meet at $q$, at which the tangent spaces coincide. Thus $\iota(\alpha)$ and $\iota(\beta)$ are tangent at $q$.
Back to the problem, when $\alpha$ is the original blue curve and $\beta$ is the original green line, then $\alpha$ and $\beta$ are tangent at $p$. Then the blue image curve $\iota(\alpha)$ and the green image circle $\iota(\beta)$ are tangent at $q=\iota(p)$. Therefore the black line that is tangent to the green circle $\iota(\beta)$ at $q$ is also tangent to the blue image curve $\iota(\alpha)$. The same applies if $\alpha$ is the original red curve and $\beta$ is the original violet line.
From the previous paragraph, it follows that if $\iota$ is conformal with respect to straight lines and circles in $\hat{\Bbb C}$ (well in fact straight lines in $\hat{\Bbb C}$ are the same as circles), then $\iota$ is conformal with respect to all differentiable curves in $\hat{\Bbb C}$. Since conformality w.r.t. straight lines and circles is known (and not difficult to establish via elementary geometry), we are done.
It is possible to establish conformality of $\iota$ solely using differential geometry as well. Let $g$ be the usual Riemannian metric on $\hat{\Bbb C}$. Show that the pullback $\iota^*g$ of $g$ under $\iota$ satisfies $(\iota^*g)_p=f(p)g_p$ for some function $f:\hat{\Bbb C}\to \Bbb R$ s.t. $f(p)>0$ for every $p\in\hat{\Bbb C}$.
